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anonymous

  • 4 years ago

calculate the integral x/[(x-a)(x-b)] when a=b

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  1. TuringTest
    • 4 years ago
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    With a=b we can write\[\int\frac x{(x-a)^2}dx\]now a little u substitution\[u=x-a\to x=u+a\to du=dx\]\[\int\frac{u+a}{u^2}du=\int u^{-1}+au^{-2}du\]Got it from here?

  2. anonymous
    • 4 years ago
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    but shouldnt we split it into partial fractions first??

  3. TuringTest
    • 4 years ago
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    I don't see a need to...

  4. TuringTest
    • 4 years ago
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    but sure, that should give the same result seems like an unnecessary pain though

  5. anonymous
    • 4 years ago
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    but i think that is what they want here and that is where i am lost

  6. TuringTest
    • 4 years ago
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    \[\int\frac x{(x-a)^2}=\int\frac A{x-a}+\frac B{(x-a)^2}dx\]\[A(x-a)+B=x\to Ax+B-Aa=x\]so we get\[A=1\]\[B-a=0\to B=a\]so our integral is\[\int\frac1{x-a}+\frac a{(x-a)^2}dx\]looks to me like we're gonna get the same answer both ways, which is a good thing :D

  7. anonymous
    • 4 years ago
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    thank you

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